Difference between revisions of "Aufgaben:Exercise 3.1Z: Influence of the Message Phase in Phase Modulation"
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− | {{quiz-Header|Buchseite= | + | |
+ | {{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM) | ||
}} | }} | ||
− | [[File:P_ID1080__Mod_Z_3_1.png|right|frame| | + | [[File:P_ID1080__Mod_Z_3_1.png|right|frame|Two PM signal waveforms]] |
− | + | We will now consider the phase modulation of diverse oscillations | |
:$$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$ | :$$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$ | ||
− | + | The source signal is represented here in normalized form with $($amplitude $1)$ , so that the phase-modulated signal can be characterised by the modulation index (or phase deviation) $η$ as follows: | |
:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$ | :$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$ | ||
− | * | + | *The signal $s_1(t)$ shown in the upper graph is characterized by the parameter values $ϕ_{\rm N} = -90^\circ$ and $η_1 = 2$ . |
− | * | + | *The frequency $f_{\rm N}$ of this sinusoidal source signal as well as the carrier frequency $f_{\rm T}$ can be determined from the signal section of duration $200 \ \rm µ s$ represented here. |
− | * | + | *The signal $s_2(t)$ possibly differs from $s_1(t)$ due to a different message phase $ϕ_{\rm N}$ and modulation index $η$. All other system parameters are unchanged from $s_1(t)$ . |
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− | '' | + | ''Hints:'' |
− | * | + | *This exercise belongs to the chapter [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]]. |
− | * | + | *Particular reference is made to the page [[Modulation_Methods/Phase_Modulation_(PM)#Signal_characteristics_of_phase_modulation|Signal characteristics of phase modulation]]. |
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Find the frequency $f_{\rm N}$ of the message signal. |
|type="{}"} | |type="{}"} | ||
$f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$ | $f_{\rm N} \ = \ $ { 5 3% } $\ \rm kHz$ | ||
− | { | + | {What is the carrier frequency $f_{\rm T}$? |
|type="{}"} | |type="{}"} | ||
$f_{\rm T} \ = \ $ { 50 3% } $\ \rm kHz$ | $f_{\rm T} \ = \ $ { 50 3% } $\ \rm kHz$ | ||
− | { | + | {What is the maximum phase deviation $ϕ_{\rm max}$ between $z(t)$ and $s(t)$? |
|type="{}"} | |type="{}"} | ||
$ϕ_{\rm max} \ = \ $ { 0.318 3% } $\ \rm rad$ | $ϕ_{\rm max} \ = \ $ { 0.318 3% } $\ \rm rad$ | ||
− | { | + | {What is the maximum time shift of the zero crossings that this phase results in? |
|type="{}"} | |type="{}"} | ||
$Δt_{\rm max} \ = \ $ { 6.37 3% } $\ \rm µ s$ | $Δt_{\rm max} \ = \ $ { 6.37 3% } $\ \rm µ s$ | ||
− | { | + | {Determine the modulation index $η_2$ for the signal $s_2(t)$. |
|type="{}"} | |type="{}"} | ||
$η_2 \ = \ $ { 2 3% } | $η_2 \ = \ $ { 2 3% } | ||
− | { | + | {What is the phase $ϕ_{\rm N2}$ of the underlying source signal $q(t)$ for $s_2(t)$ ? |
|type="{}"} | |type="{}"} | ||
$ϕ_{\rm N2} \ = \ $ { -139--131 } $\ \rm Grad$ | $ϕ_{\rm N2} \ = \ $ { -139--131 } $\ \rm Grad$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' It can be seen from the sketch that the represented section of the signal of duration $200 \ \rm µ s$ corresponds exactly to the period duration of the sinusoidal source signal. From this follows $f_{\rm N}\hspace{0.15cm}\underline{ = 5 \ \rm kHz}$. |
− | * | + | *At times $t = 0$, $t = 100 \ \rm µ s$ and $t = 200 \ \rm µ s$ , the signals $z(t)$ and $s(t)$ are synchronous in phase. |
− | *In | + | *In the first half-wave of $q(t)$ , the zero crossings of $s(t)$ come slightly earlier than those of the carrier signal $z(t)$ ⇒ positive phase. |
− | * | + | *In contrast, in the range from $t = 100 \ \rm µ s$ to $t = 200 \ \rm µ s$ , the phase $ϕ(t) < 0$. |
− | '''(2)''' | + | '''(2)''' $f_{\rm T}\hspace{0.15cm}\underline{ = 50 \ \rm kHz}$, holds, |
− | * | + | *since exactly $10$ periods can be counted in the shown section of $z(t)$ of duration $200 \ \rm µ s$ . |
− | '''(3)''' | + | '''(3)''' The maximum relative phase deviation is $ϕ_{\rm max} = η_1/(2π)\hspace{0.15cm}\underline{ ≈ 0.318}$. |
− | '''(4)''' | + | '''(4)''' Since the period of the carrier is $T_0 = 20 \ \rm µ s$ , we obtain $Δt_{\rm max} = ϕ_{\rm max} ·T_0\hspace{0.15cm}\underline{ ≈ 6.37 \ \rm µ s}$. |
− | '''(5)''' | + | '''(5)''' The maximum phase deviation (shift in the zero intercepts) is exactly the same for $s_2(t)$ as for $s_1(t)$. |
− | * | + | *From this, we can conclude that $η_2 = η_1\hspace{0.15cm}\underline{ = 2}$ . |
− | '''(6)''' | + | '''(6)''' The signal $s_2(t)$ is shifted to the right by $25 \ \rm µ s$ compared to $s_1(t)$ . Therefore, the same must be true for the source signals: |
:$$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$ | :$$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$ | ||
− | * | + | *This corresponds to the phase position $ϕ_{\rm N2}\hspace{0.15cm}\underline{ = -135^\circ}$. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category:Modulation Methods: Exercises|^3.1 | + | [[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]] |
Latest revision as of 16:54, 9 April 2022
We will now consider the phase modulation of diverse oscillations
- $$ q(t) = \cos(\omega_{\rm N} \cdot t + \phi_{\rm N})\hspace{0.05cm}.$$
The source signal is represented here in normalized form with $($amplitude $1)$ , so that the phase-modulated signal can be characterised by the modulation index (or phase deviation) $η$ as follows:
- $$s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + \eta \cdot q(t) \big]\hspace{0.05cm}.$$
- The signal $s_1(t)$ shown in the upper graph is characterized by the parameter values $ϕ_{\rm N} = -90^\circ$ and $η_1 = 2$ .
- The frequency $f_{\rm N}$ of this sinusoidal source signal as well as the carrier frequency $f_{\rm T}$ can be determined from the signal section of duration $200 \ \rm µ s$ represented here.
- The signal $s_2(t)$ possibly differs from $s_1(t)$ due to a different message phase $ϕ_{\rm N}$ and modulation index $η$. All other system parameters are unchanged from $s_1(t)$ .
Hints:
- This exercise belongs to the chapter Phase Modulation.
- Particular reference is made to the page Signal characteristics of phase modulation.
Questions
Solution
- At times $t = 0$, $t = 100 \ \rm µ s$ and $t = 200 \ \rm µ s$ , the signals $z(t)$ and $s(t)$ are synchronous in phase.
- In the first half-wave of $q(t)$ , the zero crossings of $s(t)$ come slightly earlier than those of the carrier signal $z(t)$ ⇒ positive phase.
- In contrast, in the range from $t = 100 \ \rm µ s$ to $t = 200 \ \rm µ s$ , the phase $ϕ(t) < 0$.
(2) $f_{\rm T}\hspace{0.15cm}\underline{ = 50 \ \rm kHz}$, holds,
- since exactly $10$ periods can be counted in the shown section of $z(t)$ of duration $200 \ \rm µ s$ .
(3) The maximum relative phase deviation is $ϕ_{\rm max} = η_1/(2π)\hspace{0.15cm}\underline{ ≈ 0.318}$.
(4) Since the period of the carrier is $T_0 = 20 \ \rm µ s$ , we obtain $Δt_{\rm max} = ϕ_{\rm max} ·T_0\hspace{0.15cm}\underline{ ≈ 6.37 \ \rm µ s}$.
(5) The maximum phase deviation (shift in the zero intercepts) is exactly the same for $s_2(t)$ as for $s_1(t)$.
- From this, we can conclude that $η_2 = η_1\hspace{0.15cm}\underline{ = 2}$ .
(6) The signal $s_2(t)$ is shifted to the right by $25 \ \rm µ s$ compared to $s_1(t)$ . Therefore, the same must be true for the source signals:
- $$ q_2(t) = q_1(t - 25\,{\rm \mu s}) = \cos \hspace{-0.1cm} \big[2 \pi f_{\rm N} (t - 25\,{\rm \mu s}) \big ] = \cos (\omega_{\rm N} \cdot t - 0.75 \cdot \pi)\hspace{0.05cm}.$$
- This corresponds to the phase position $ϕ_{\rm N2}\hspace{0.15cm}\underline{ = -135^\circ}$.